Problem: Let $f(x)=\tan(x)$. Find $f'\left(\dfrac{\pi}{6}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3}{4}$ (Choice B) B $4$ (Choice C) C $\dfrac{1}{4}$ (Choice D) D $\dfrac{4}{3}$
Answer: Let's first find $f'(x)$. Then, we can evaluate it at $x=\dfrac{\pi}{6}$. Recall that the derivative of $\tan(x)$ is $\dfrac{1}{\cos^2(x)}$, or $\sec^2(x)$. [Is there a way to know this without memorizing?] So $f'(x)=\dfrac{1}{\cos^2(x)}$. Now let's plug $x={\dfrac{\pi}{6}}$ into $f'$ : $\begin{aligned} &\phantom{=}f'\left({\dfrac{\pi}{6}}\right) \\\\ &=\dfrac{1}{\cos^2\left({\dfrac{\pi}{6}}\right)} \\\\ &=\dfrac{1}{\left(\dfrac{\sqrt 3}{2}\right)^2} \\\\ &=\dfrac{1}{\left(\dfrac{3}{4}\right)} \\\\ &=\dfrac{4}{3} \end{aligned}$ In conclusion, $f'\left(\dfrac{\pi}{6}\right)=\dfrac{4}{3}$.